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1. Homework Statement
(The following is taken from Sears and Zemansky’s University physics with modern physics, thirteenth edition by Young Freedman. Chapter 3 bridging problem: Launching up an incline at page 95.
You fire a ball with an initial speed ##v_0## at an angle Φ above the surface of an incline, which is itself inclined at an angle θ above the horizontal (see the above image). (a) Find the distance measured along the incline, from the launch point to the point when the ball strikes the incline. (b) What angle Φ gives the maximum range, measured along the incline? Ignore air resistance.
Identify and set up:
- Since there’s no air resistance, this is a problem in projectile motion. The goal is to find the point where the ball’s parabolic trajectory intersects the incline.
- Choose the x- and y-axes and the position of the origin. When in doubt, use the suggestions given in Problem-Solving Strategy 3.1 in Section 3.3
- In the projectile equations from Section 3.3, the launch angle α0 is measured from horizontal. What is this angle in terms of θ and Φ? What are the initial x- and y-components of the ball’s initial velocity?
- You’ll need to write and equation that relates x and y for points along the incline. What is this equation? (This takes just geometry and trigonometry, not physics.)
Execute:
- Write the equations for the x-coordinate and y-coordinate of the ball as functions of time t.
- When the ball hits the incline, x and y are related by the equation that you found in step 4. Based on this, at what time t does the ball hit the incline?
- Based on your answer from Execute: step 2, at what coordinates x and y does the ball land on the incline? How far is this point from the launch point?
- What value of Φ gives the maximum distance from the launch point to the landing point? (Use your knowledge of calculus.)
Evaluate:
- Check your answers for the case θ = 0, which corresponds to the incline being horizontal rather than tilted. (You already know the answers for this case. Do you know why?)
(That is all from the book. From now on it will be me speaking. I am not a native English speaker and wasn’t taught math in English so I hope you will forgive me if I don’t manage to explain my thoughts too well.
I should also mention that I study philosophy, but are trying to learn some physics by myself on the side, so there might be some holes in my knowledge.)
From the same book I know that the answers are (a) ##R = \frac{2v_0{}^2}{g} \frac{cos(\theta + \phi)sin(\phi)}{cos^2(\theta)}## and (b) ##\phi = 45^{\circ}- \frac{\theta}{2}##
I already got a friend to help me with (a) but I can’t seem to understand (b). I would really appreciate some help with it.
Homework Equations
##\frac{d}{dx}sin(ax) = acos(ax)##
##\frac{d}{dx}cos(ax) = -asin(ax)##
##sin(a+b)= sin(a)cos(b)+cos(a)sin(b)##
##cos(a+b) = cos(a)cos(b)- sin(a)sin(b)##
The Attempt at a Solution
So my goal is to find the value of Φ which gives the highest value of R.
My initial thought is to take the derivative of R, ##\frac{dR}{d\phi}##, and then say ##\frac{dR}{d\phi} = 0## to find the points where R is highest/lowest.
First I use ##cos(a+b) = cos(a)cos(b)- sin(a)sin(b)## so I get ##R = \frac{2v_0{}^2}{g} \frac{sin(\phi)(cos(\phi)cos(\theta)- sin(\phi)sin(\theta))}{cos^2\theta}##
Then I take the derivative using ##\frac{d}{dx}sin(ax) = acos(ax)## and ##\frac{d}{dx}cos(ax) = -asin(ax)##
I get ##\frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{cos(\phi)(-sin(\phi)cos(\theta)- cos(\phi)sin(\theta))}{cos^2\theta}
\leftrightarrow \frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{-cos(\phi)(sin(\phi)cos(\theta)+ cos(\phi)sin(\theta))}{cos^2\theta}##
And at last, I use ##sin(a+b)= sin(a)cos(b)+cos(a)sin(b)##
To get: ##\frac{dR}{d\phi} = \frac{2v_0{}^2}{g} \frac{-cos(\phi)sin(\phi+\theta)}{cos^2\theta}##
This seems to give two solutions. Either:
##-cos(\phi) = 0## or ##sin(\phi+\theta) = 0##
Those solutions gives the answers ##\phi = 90^{\circ}## and ##\phi = -\theta##
If I substitute those values for ##\phi## in R I get:
##R = \frac{2v_0{}^2}{g} \frac{-sin(\theta)}{cos^2(\theta)}## and ##R = \frac{2v_0{}^2}{g} \frac{sin(\theta)cos(2\theta)}{cos^2\theta}##
For an ordinary maximization problem, this would be the part where I test whether the values of Φ for ##\frac{dR}{d\phi} = 0## gives any maximum.
Before doing that, there are a few implicit assumptions, which I would like to try to make explicit.
First of, if ##v_0{}^2 = 0##, the values of Φ and θ won’t matter since R will be equal to 0. I will therefore assume that ##v_0{}^2 \neq 0## and since ##v_0{}^2 ## can’t be negative either, it means that ##\frac{2v_0{}^2}{g}## can be treated as a positive constant.
Secondly, ##90^{\circ}>\theta >-90^{\circ}##. This is because ##cos^2(\theta) \neq 0## and because ##90^{\circ} >\phi + \theta>-90^{\circ}##. If ##\phi + \theta<-90^{\circ}## or ##90^{\circ} <\phi + \theta##, the ball will never hit the incline due to gravity and the equation will have no meaning.
Lastly, based on this, ##cos(\theta + \phi)sin(\phi)## seems to be the only part of the equation which matters, since ##\frac{2v_0{}^2}{g}## can be treated as a positive constant as shown above and ## cos^2(\theta)## is irrelevant as long as ## cos^2(\theta) \neq 0## Therefore, as long as ##cos(\theta + \phi)sin(\phi)## has maximum value, R will have maximum value.
Based on these assumptions, I calculated the value of R for ##\theta = -10##, ##\theta = 0## and ##\theta = 10## For each value of θ I calculated the value of R if ##\phi = 89^{\circ}##, if ##\phi = 90^{\circ}##, if ##\phi = 91^{\circ}##, if ##\phi = \theta -1##, if ##\phi = \theta ## and if ##\phi = \theta +1##.
Based on the results, neither ##\phi = 90^{\circ}## or ##\phi = -\theta## appear to be a maximum.
Of course, I already knew this since I know the answer is ##\phi = 45^{\circ}- \frac{\theta}{2}## but I completed these steps anyway, hoping to get some clue as to what to do next. Sadly that didn’t happen, and know I don’t really know where to go from here. Any help would be much appreciated.